Question

after falling from rest from a height of 30 m, a 0.50 kg ball rebounds upward, reaching a height of 20 m. if the contact between bam and ground lasted 2.0 ms, what average force was exerted on the ball?

Answer 1

Solution: The mass of the ball is m = 0.50 kg

The ball falls from the rest from the height of 30 m. Thus its velocity (initial) just before impact with the ground can be calculated by,

v_i² = u² + 2*a*y

We have u = 0 m/s (ball falls from the rest), a = g = -9.8 m/s² and y = -30 m (negative displacement)

v_i² = (0m/s)² + 2*( -9.8 m/s²)*(-30m)

v_i² = 588 m²

v_i = - 24.25 m/s

We take negative value for the velocity as the velocity is in negative direction

Thus v_i = - 24.25 m/s is the initial velocity of the ball just before the impact.

Now the ball bounces back and reaches the maximum height of 20 m.

Let v_a be the velocity of the ball just after the impact. At the maximum height the velocity of the ball is zero, that is v_f = 0 m/s. The height reached is y’ = 20 m

v_f² = v_a² + 2*a*y

(0 m/s)² = v_a² + 2*( -9.8 m/s²)*(20m)

v_a² = 392 m²

v_a = 19.80 m/s

Now from the impulse momentum theorem we have,

J = change in momentum = Δp = (final momentum – initial momentum)

J = m*v_a – m*v_i

J = 0.5kg*(19.80m/s) – 0.5kg*(-24.25m/s)

J = 22.025 kg^.m/s

During the impact, the contact between the ball and ground lasted for Δt = 2.0 ms = 2.0*10^-3 s

Thus the average force on the ball exerted by the ground is given by,

F_average = J/Δt

F_average = (22.025 kg.m/s)/( 2.0*10^-3 s)

F_average = 11013 N or

F_average = 1.10*10⁴ N

Hence the answer is 1.10*10⁴ N